# What is d/(d theta) (sec^2 4theta) ?

Jul 23, 2016

$\frac{d}{d \theta} \left({\sec}^{2} 4 \theta\right) = 8 {\sec}^{2} 4 \theta \tan 4 \theta$

#### Explanation:

$\frac{d}{d \theta} \left({\sec}^{2} 4 \theta\right) = 2 \sec 4 \theta \cdot \frac{d}{d \theta} \left(\sec 4 \theta\right)$ (Power rule and Chain rule)

$= 2 \sec 4 \theta \cdot \tan 4 \theta \sec 4 \theta \cdot 4$ (Standard differential and Chain rule)

$= 8 {\sec}^{2} 4 \theta \tan 4 \theta$

Apr 4, 2018

$8 {\sec}^{2} \setminus 4 \theta \tan 4 \theta$

#### Explanation:

Given: $\frac{d}{d \setminus \theta} \left({\sec}^{2} \setminus 4 \theta\right)$

We use the chain rule, which states that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

In this case, it's:

$\frac{\mathrm{dy}}{d \setminus \theta} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{d \setminus \theta}$

And so, let $u = 4 \theta$, then $\mathrm{du} = 4 \setminus d \setminus \theta , \frac{\mathrm{du}}{d \setminus \theta} = 4$.

Also, $y = {\sec}^{2} u$, then $\frac{\mathrm{dy}}{\mathrm{du}} = 2 {\sec}^{2} u \tan u$.

Therefore, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {\sec}^{2} u \tan u \cdot 4$

$= 8 {\sec}^{2} u \tan u$

Reversing back our substitution, we get,

$= 8 {\sec}^{2} \setminus 4 \theta \tan 4 \theta$